which statement is true about the discontinuities of the function f(x)? f(x) = (x + 1)/(6x² - 7x - 3) there…

Explanation:

Step1: Factor the denominator

Factor $6x^{2}-7x - 3$. We need two numbers $a$ and $b$ such that $a\times b=6\times(- 3)=-18$ and $a + b=-7$. The numbers are $-9$ and $2$. So, $6x^{2}-7x - 3=6x^{2}-9x+2x - 3 = 3x(2x - 3)+1(2x - 3)=(2x - 3)(3x + 1)$. Then $f(x)=\frac{x + 1}{(2x - 3)(3x+1)}$.

Step2: Find the vertical - asymptotes

Vertical asymptotes occur at the values of $x$ that make the denominator equal to zero while the numerator is non - zero. Set the denominator equal to zero: $(2x - 3)(3x + 1)=0$. Solving $2x-3 = 0$ gives $x=\frac{3}{2}$, and solving $3x + 1=0$ gives $x=-\frac{1}{3}$. Since the numerator $x + 1$ is non - zero at $x=\frac{3}{2}$ and $x=-\frac{1}{3}$ (when $x=\frac{3}{2}$, $x + 1=\frac{3}{2}+1=\frac{5}{2}\neq0$; when $x=-\frac{1}{3}$, $x + 1=-\frac{1}{3}+1=\frac{2}{3}\neq0$), the vertical asymptotes are at $x=\frac{3}{2}$ and $x=-\frac{1}{3}$.

Answer:

There are asymptotes at $x=\frac{3}{2}$ and $x =-\frac{1}{3}$.